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Euclidean, Spherical and Hyperbolic Shadows
Ryan Hoban
Consider this familiar related rates problem from freshman calculus:
([7] pg 181) A street light is mounted at the top of a 15-ft tall
pole. A man 6 feet tall walks away from the pole with a speed of
5 ft/sec along a straight path. How fast is the tip of his shadow
moving when he is 40 ft from the pole?
The standard solution uses similar triangles, a feature uniquely Euclidean,
but it can be solved another way using intersections of lines. The same
question can then be posed and solved in any geometry where we can compute
intersections of lines. Here we solve the problem for an individual walking in
the elliptic or hyperbolic plane. Working in the other geometries illustrates
key differences between the flat geometry of Euclidean space and these other
constant curvature, two dimensional geometries.
The setup
Let the individual have fixed height h and the lamppost have fixed height
l (of course assuming h < l). Let s be the length of the shadow and d
the distance from the base of the lamppost to the individual’s feet. A good
calculus student will draw the picture in Figure 1.
Solving the problem requires that we write s as a function of d. In the
Euclidean plane, the ground is a straight line, as is the light ray from the lamp
1
h
l
sd
H
T
F
L
Figure 1: The setup in the Euclidean plane.
through the top of the individual’s head. In the other two plane geometries we
draw similar schematic pictures. We assume that the ground is flat and the
light ray travels along a straight path, that is both are distance minimizing
curves in the appropriate geometry. Such a curve is called a geodesic.
Figure 2 is the setup in spherical geometry, shown both on the sphere and
after stereographic projection to the plane. Figure 3 gives the setup in two
models of hyperbolic geometry, the Poincare unit disk and the upper half
plane ([1] describes these models).
We compute s as a function of d in each geometry. The result is a fantastic
illustration of the effect curvature has on these, and nicely illustrates the
duality between spherical and hyperbolic geometry.
Before reading the remainder of this article, the reader might enjoy ex-
amining figures 2 and 3 to get some intuition for what to expect from the
shadow in each geometry. If imagination is not enough, the animations avail-
able online at [3] should convince you that the shadow on the sphere is peri-
odic, while in the hyperbolic plane the shadow length grows to infinity very
2
l
d
h
L
H
F=T
l
dh
s
H
L
F
T
Figure 2: The setup in spherical geometry.
rapidly. The computations are a bit complicated, but the pictures make the
complicated formulas believable.
Euclidean Shadows
To solve the problem, we are led to write down the shadow length s as a
function of Stickman’s distance d from the base of the light post. We assume
that the ground lies along the x-axis and the light post along the y-axis.
This enables us to write the equation of the light ray from the light source
through the top of Stickman’s head.
y =(h−ld
)x+ l
The tip of the shadow is the x-intercept of this line,(ldl−h , 0
). The shadow
length is simply the distance from the shadow tip to Stickman’s feet, i.e.
3
Figure 3: The setup in hyperbolic geometry.
s = dist(feet,shadowtip)
= dist
((d, 0) ,
(ld
l − h, 0
))= d
(h
l − h
) .
We see that shadow length is directly proportional to the distance walked.
The shadow becomes arbitrarily long, but always has finite length. The
problem can be finished by differentiating to obtain the rate of change of
shadow length and adding it to the speed of the individual moving on the
ground.
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Spherical Shadows
Let’s now compute s as a function of d assuming Stickman walks in spherical
geometry. The standard model for spherical geometry is the unit sphere
centered at the origin in R3. Here geodesics are great circles, that is the
intersection of the unit sphere with linear subspaces of R3. Points in spherical
geometry are represented by unit vectors in R3 and the Euclidean metric on
R3 induces a metric on the sphere. The spherical distance between two points
is simply the Euclidean arc length along the great circular arc connecting
these points. Let SD(u, v) denote the spherical distance between unit vectors
u and v, which is related to the dot product by
cos (SD(u, v)) = u · v.
We assume that the ground is the sphere’s equator and the light post lies
along a meridian. Observe that the ground then has finite length 2π and the
farthest the light can be away from the ground is π2. Even before working
out the computation, Figure 4 makes it clear that there is a key difference
between l = π2
and l < π2. Specifically, when l = π
2, the light ray intersects
the top of Stickman’s head at a right angle, then runs along his body, so
no shadow is produced. He is always standing directly under the light, no
matter where he walks. Figure 4 also suggests that if l < π2, the shadow
length will be periodic, not surprising as the whole geometry is finite.
Now for some coordinates. Suppose the base of the lamp is at
1
0
0
.
If the light post lies along the meridian through that point and has height l,
then the light source is at L =
cos(l)
0
sin(l)
. After walking distance d from the
base of the light post then Stickman’s feet are at F =
cos(d)
sin(d)
0
. Assuming
5
l
d
h
L
H
F=T
l
d
hs
LH
FT
Figure 4: Stickman on the sphere.
Stickman has height h, the top of his head is at H =
cos(d) cos(h)
sin(d) cos(h)
sin(h)
.
We can now determine the geodesic light ray which forms the shadow.
The light source L and the top of Stickman’s head H are unit vectors in
R3 which determine a unique plane through the origin. The intersection of
that plane with the unit sphere is the geodesic light ray. The shadow tip
is the intersection of this plane with the equator. The tip of the shadow,
T , is determined by the fact that it lies along the intersection of the plane
containing L and H and the equatorial plane - and the fact that it is a unit
vector. Its coordinates are found by solving simultaneously (L×H) · T = 0,
‖T‖ = 1 and z = 0,
This system has 2 antipodal solutions. With the aid of a computer algebra
system we obtain:
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T =
cos(d) cos(h) sin(l)−sin(h) cos(l)√
−2 cos(d) sin(h) cos(h) sin(l) cos(l)+sin2(h) cos2(l)+cos2(h) sin2(l)sin(d) cos(h) sin(l)√
−2 cos(d) sin(h) cos(h) sin(l) cos(l)+sin2(h) cos2(l)+cos2(h) sin2(l)
0
.
The shadow length s is the spherical distance from the Stickman’s feet F
to the tip of the shadow T . We find that
cos(s) = F · T
=cos(h) sin(l)− cos(d) sin(h) cos(l)√
−2 cos(d) sin(h) cos(h) sin(l) cos(l) + sin2(h) cos2(l) + cos2(h) sin2(l)
This is indeed complicated and scary, but note that if l = π2
the entire
expression simplifies to cos(s) = 1. In this case shadow length is identically
0, as expected. Figure 5 graphs s as a function of d for fixed light height
l = π3
and various choices for h. The shadow length is periodic. What is less
obvious is that as h→ l the relationship between s and d approaches a linear
relationship, though not uniformly.
Π 2 ΠDist Walked
-Π
2
0
Π
2
Shadow Length
Figure 5: Spherical shadows lengths.
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Hyperbolic Shadows
What if Stickman is walking in the hyperbolic plane? For readers unfamiliar
with this geometry there are a myriad of great references. Specifically [1],
[4] and [5] are excellent introductions. The many different models of the
hyperbolic plane often present a challenge when first learning the subject
as Euclidean and spherical geometry each have a single standard model.
The relationships between different models however is key to understanding
hyperbolic geometry as each model has its own advantages & disadvantages.
Anderson ([1]) does a particularly good job relating the various models.
For computational ease, we work in the upper half plane model. This is
the upper half of the complex plane {x + iy : y > 0}. The geodesics are
either vertical rays or circular arcs intersecting the real axis at a right angle.
z1
w1
z1*
w1*=¥
z2
w2
z2* w2
*
Figure 6: Typical geodesics in the upper half plane.
For the shadow problem, we need a formula for distances. Given distinct
points z and w in the upper half plane we connect these with a geodesic
segment. If Re(z) = Re(w) this is the vertical line segment connecting z and
w, otherwise it is a circular arc. Extend this geodesic segment to a complete
geodesic and observe that this intersects R ∪ {∞} twice. The endpoints are
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called ideal points and we denote them by z∗ and w∗, so that z is between
z∗ and w. If the geodesic is a vertical line, one of the ideal endpoints is ∞(see figure 6). Let HD(z, w) denote the hyperbolic distance between z and
w, which by definition is
HD(z, w) =
ln(
(z−w∗)(w−z∗)(w∗−w)(z∗−z)
)if neither z∗ nor w∗ is ∞
ln(
Im(z)Im(w)
)if z∗ =∞
ln(
Im(w)Im(z)
)if w∗ =∞
(1)
.
To calculate s as a function of d, assume the ground is the positive imag-
inary axis and the base of the light post is at i. If the light post is per-
pendicular to the ground then it lies along the the unit circle. The lamp
then has a distance l from i, and this is the point L = tanh(l) + i sech(l).
After walking along the ground a distance d, Stickman’s feet are at F = ied.
Assuming he is standing upright means his body lies along the circular arc
centered at the origin of radius ed. Since his height is h, the top of his head
is H = ed(tanh(h) + i sech(h)). The reader is encouraged to use 1 to check
that HD(i, F ) = d, HD(i, L) = l and HD(F,H) = h. In fact an excellent
exercise would be to derive these coordinates yourself (Hint: The lamp post
has its base at i and the ideal points of that geodesic are −1 and 1).
Let T denote the tip of the shadow, the intersection of the geodesic light
ray with the ground. The light ray is the geodesic arc through L and H,
which we can explicitly compute. The fact that it intersects the real axis at
a right angle means that the circle is centered at some point on the real axis,
say x0. If r is its radius, then this geodesic satisfies the equation
(x− x0)2 + y2 = r2.
Substituting in the real and imaginary parts of L and H, we obtain a
system with unknowns x0 and r
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h
l
d
s
L=tanhHlL+i sechHlL
H=edHtanhHhL+i sechHhLL
F=ied
T
Figure 7: The setup in the upper half plane.
(tanh(l)− x0)2 + ( sech(l))2 = r2
(ed tanh(h)− x0)2 + (ed sech(h))2 = r2,
which we can solve for x0 and r. With the aid of a computer algebra system
we obtain
T = i
√ed(tanh(h)−ed tanh(l))ed tanh(h)−tanh(l)
.
To compute s we simply need the distance from F to T . Since these
points both lie on the imaginary axis the distance is
s = ln
(e−d
√ed (tanh(h)− ed tanh(l))
ed tanh(h)− tanh(l)
)(2)
.
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We now have the desired relationship between s and d. This looks com-
plicated but a brief analysis confirms what we might already suspect. Note
that s is undefined if the numerator or denominator of the radicand becomes
zero. This occurs when
d = d0 = ± ln(
tanh(l)tanhh
).
This has a wonderful interpretation. After walking a finite distance d0 in
either direction Stickman’s shadow has infinite length! Figure 8 illustrates
this. When d = d0 the hyperbolic ray forming the shadow becomes a vertical
line which does not intersect the imaginary axis (ground). When d = −d0the ray will becomes a circular arc whose endpoint is the origin which also
does not intersect the positive imaginary axis. What a sharp contrast with
the Euclidean case, where the shadow always is finite.
ie-d0
tanhHlL+i sechHlL
tanhHlL+i tanhHlL cschHhL
ied0
Figure 8: A finite distance from the lamp, Stickman’s shadow becomes infi-
nite.
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Duality
We conclude with a comparison of these seemingly unrelated geometries.
Figure 9 graphs the solutions for each geometry for fixed l and h satisfying
0 < h < l < π2.
d=d0
hyperbolic
Euclideanspherical
Π 2 Πd
-Π
2
0
Π
2
s
Figure 9: Shadow length s as a function of d in all three geometries.
Recall that the solution in the hyperbolic case was
s = ln
(e−d√
ed(tanh(h)−ed tanh(l))ed tanh(h)−tanh(l)
).
Taking cosh of both sides and then applying a handful of hyperbolic
trigonometric identities, we rewrite this as
cosh(s) = cosh(d) sinh(h) cosh(l)−cosh(h) sinh(l)√cosh2(l) sinh2(h)−2 cosh(d) cosh(h) cosh(l) sinh(h) sinh(l)+cosh2(h) sinh2(l)
.
Compare this with the solution obtained in the spherical case which was
cos(s) = cos(h) sin(l)−cos(d) sin(h) cos(l)√cos2(l) sin2(h)−2 cos(d) sin(h) cos(h) sin(l) cos(l)+cos2(h) sin2(l)
.
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These formulas make dramatically clear the duality between spherical and
hyperbolic geometry. In particular since cosh(iθ) = cos(θ) and sinh(iθ) =
i sin(θ), we can transform the hyperbolic relationship into the spherical one
by multiplying all variables by −i. It can be shown that if all variables
are small, the solutions are approximately Euclidean. In other words, the
solutions agree up to first order. Stickman should be assured that if his
journey is in a direction with curvature close to zero and he doesn’t walk too
far, his shadow will behave approximately as in flat Euclidean space.
Final Remark
My own motivation for this came after answering the original question for a
calculus class, then realizing that our solution using similar triangles was only
applicable in Euclidean geometry. When teaching an undergraduate course
one should pay attention when unnecessarily making implicit use of Euclidean
geometry. Many classical problems from pre-calculus and calculus use facts
about triangles or Euclidean trigonometry, and you can answer the same
questions in spherical or hyperbolic trigonometry. I don’t advise confusing
students first learning calculus with these issues, but exciting investigations
are there to be found by changing the geometry.
Acknowledgment I am grateful to the referees for many helpful com-
ments on this paper. I would like to thank William Goldman for all of his
patience and support, as well as Anton Lukyanenko, Stefan Mendez-Diez,
Andy Sanders and Robin Shutinya for helpful feedback on this project. I am
grateful for support from the University of Maryland Mathematics Depart-
ment, and the National Science Foundation grant of Goldman, DMS070781.
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References
[1] J. Anderson, Hyperbolic Geometry, Springer Verlag, London UK, 2005
[2] A. Baragar, A Survey of Classical and Modern Geometries, Prentice
Hall, 2001
[3] R. Hoban, University of Maryland Mathematics Department (2008)
http://www.math.umd.edu/∼rfhoban/Shadows.html
[4] Mumford, Series, & Wright, Indra’s Pearls: The Vision of Felix Klein,
Cambridge University Press, Cambridge UK,2002
[5] M. Helena Noronha, Euclidean and Non-Euclidean Geometries, Prentice
Hall, Upper Saddle River NJ, 2002
[6] J.G. Ratcliffe, Foundations of Hyperbolic Manifolds, Graduate Texts in
Mathematics vol. 149, Springer-Verlag NY, 1994
[7] J. Stewart, Single Variable Calculus Thomson Brooks/Cole, The Thom-
son Corporation, Belmont CA, 2005
[8] W. Thurston, Three Dimensional geometry and topology, Vol. 1 (S.
Levy ed.) , Princeton Mathematical Series, Princeton University Press,
Princeton NJ, 1997
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